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So, recently while I was revisiting rotational dynamics, I came across this question: A rod AB of length L stands vertically on a horizontal floor and its end B is at height L. Now the end B starts falling down because of some slight disturbance and end A does not slip. Calculate the speed with which the end B hits the ground.

See, this question can be easily done by applying mechanical energy conservation. Since the end A is not moving there must be enough friction there, even then we can easily apply energy conservation because the point where friction acts (end A) does not displace, hence work done by friction is zero. But why can't we apply angular momentum conservation about the centre of mass of that rod as there won't be any net torque about CM? (As I said, the point where friction acts doesn't move so its work is zero, and in case of gravity, the torque due to weight will be zero as we are considering our axis to pass through CM.)

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  • $\begingroup$ $ PE = m g L/2; KE = \frac12 m v^2 + \frac12 I \omega^2; $ $\endgroup$
    – Narasimham
    Commented Nov 26, 2019 at 18:31
  • $\begingroup$ The normal force at the rod contact with the ground produces a torque about the cm. $\endgroup$
    – Bill Watts
    Commented Nov 26, 2019 at 18:35
  • $\begingroup$ @Narasimham Thanks but I am asking why can't we solve it by using angular conservation as friction does no work $\endgroup$
    – jatin Goyal
    Commented Nov 26, 2019 at 18:35
  • $\begingroup$ Torque is $r\times F$, so the friction force produces a torque also even if it does no work. But if your cm is your relative stationary point, the contact with the floor is moving relative to that. $\endgroup$
    – Bill Watts
    Commented Nov 26, 2019 at 18:43
  • $\begingroup$ @BillWatts Thanks sir, but i have some query related to this.. 1) if we have a rod and a torque acts on it at some random point A , now if we consider another point B on that same rod so can we apply angular momentum conservation given that the point A is at rest throughout from B's frame of reference $\endgroup$
    – jatin Goyal
    Commented Nov 26, 2019 at 18:47

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Although the static friction does no work it does apply a force to the bottom of the rod (which accelerates the center of mass horizontally), and a torque. The normal force also produces a torque relative to te center of mass. Both of these torques change as the rod falls, and cause changes in the angular momentum. (The angular acceleration is a variable, and the time of fall is strongly dependent on the initial conditions.) (There is likely to be an angle at which the normal force is no longer sufficient to maintain static friction. At that point the bottom starts to slip and things get even more complicated.)

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  • $\begingroup$ Thanks sir, but i have some query related to this.. 1) if we have a rod and a torque acts on it at some random point A , now if we consider another point B on that same rod so can we apply angular momentum conservation given that the point A is at rest throughout from B's frame of reference $\endgroup$
    – jatin Goyal
    Commented Nov 26, 2019 at 18:53
  • $\begingroup$ In a static situation, the sum of torques about any chosen point should be zero. If rotation is allowed around a fixed point, you wold normally take torques about that point. (In the case of your falling rod, torque about the lower end is caused by gravity acting at the center of gravity.) (In a uniform field, the center of gravity is at the center of mass.) In general, the vector sum of all forces acting on an object determines the acceleration of the center of mass, and the sum of torques about the center of mass determines the angular acceleration around the center of mass. $\endgroup$
    – R.W. Bird
    Commented Nov 27, 2019 at 17:39

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